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COMMENTS.    By  S.  W.  Robinsoo.  C.E 


STRENGTH  OF  BEAMS 


TRANSVERSE  LOADS. 


WILLIAM  ALLAN,  A.M.,  L.L.D., 

i\ 

LATE   PRINCIPAL   OF   McDoNOCte   SCHOOL,    MARYLAND. 

AUTHOR    OF    "CHANCELLORSVILLE,"    "JACKSON'S 

VALLEY  CAMPAIGN,"  "  THE  ARMY  OF  NORTHERN 

VIRGINIA   IN    1862,"    ETC.,    ETC. 

SECOND   EDITION,  REVISED. 


NEW   YORK: 

D.    VAN  NOSTRAND   COMPANY, 

23  MURRAY  AND  27  WARREN  STREET. 

1893. 


PREFACE. 


The  usual  discussion  of  stress  in  beams  loaded 
transversely,  involving  as  it  does  the  calculus,  is 
unintelligible  to  that  large  class  of  builders,  and 
others,  whose  mathematical  training  has  not  exten- 
ded beyond  the  elements  of  conic-sections.  Yet,  the 
most  important  cases  may  be  explained  without 
using  the  higher  mathematics. 

The  aim  of  the  following  pages  is  to  put  into 
brief  and  convenient  shape,  without  resorting  to 
the  higher  mathematics,  the  discussion  of  the  most 
important  and  common  cases  of  horizontal  beams 
under  vertical  loads.  In  beams  of  rectangular 
cross-section  the  formulae  given  are  exact  under  the 
assumptions  generally  made  in  regard  to  the  laws 
of  elasticity.  In  flanged  beams  the  approximations 
made  use  of  are  those  adopted  in  ordinary  practice. 

A  graphic  method  of  determining  the  amount 
of  resistance  is  added,  which  is  applicable  to  all 
cross  sections,  and  which,  if  used  with  ordinary 
skill  and  care,  will  give  results  exact  enough  for 
every  purpose.  This  method  is  so  simple,  and  of 
such  general  application,  that  it  commends  itself  to 


6 


practical  builders.  It  is  easily  understood,  and  re- 
quires no  other  appliances  but  those  in  the  hands 
of  every  draughtsman.  In  the  case  of  unsymmetrical 
cross-sections,  (as  for  instance,  iron  rails)  it  affords 
an  easy  means  of  obtaining  sufficiently  accurate 
results,  without  resorting  to  the  tedious  calculations 
otherwise  necessary. 

This  discussion  (in  which  there  is  no  attempt  at 
originality)  was  prepared  for  the  use  of  the  inter- 
mediate classes  in  Engineering,  at  Washing- 
ton and  Lee  University.  Having  been  found 
serviceable  to  them,  both  during  their  college 
career  and  in  subsequent  practice,  I  am  induced 
to  hope  that  it  may  save  time  and  labor  to  those 
engaged  in  the  same  pursuits. 

In  prepanng  the  manuscript  for  the  press,  and 
correcting  the  proofs,  I  have  been  much  indebted 
to  one  of  my  former  pupils,  Mr.  Julius  Krutt- 
schnitt,  C.  E.,  now  Instructor  in  the  McDonogh 
School. 

AUGUST,  1875. 


STRENGTH  OF  BEAMS 
UNDER  TRANSVERSE   LOADS. 


IN  the  following  discussion  the  ordi- 
nary cases  of  loaded  beams  are  treated 
without  resorting  to  the  higher  mathema- 
tics. It  is  an  attempt  to  compile  from 
various  sources  the  simplest  methods  of 
treating  such  cases  as  arise  most  frequently 
in  practice. 

Transverse  stress  is  produced  by  a  load 
applied  to  a  beam  in  a  direction  perpen- 
dicular, or  inclined,  to  its  length.  A  D 
(Fig.  1)  is  a  beam  subjected  to  such  a 
stress. 

In  this  kind  of  stress  a  compression  of 
the  particles  or  fibres  on  one  side  of  the 


A 
0 

B 


FIG.  1. 

beam  and  an  extension  of  those  on  the 
other,  are  produced.  In  consequence  of  this 
the  beam  bends.  Experiment  shows  that 
the  amount  of  compression  on  the  one  side, 
and  of  extension  on  the  other,  diminishes 
as  we  go  inwards  from  the  top  or  bottom 
towards  the  centre,  and  at  some  intermedi- 
ate plane,  O  O',  becomes  zero.  The  fibres 
at  this  plane  being  neither  lengthened  nor 
shortened,  it  is  called  the  neutral  plane,  and 
its  intersection  by  the  plane  of  vertical 
section  is  called  the  neutral  line  or  axis. 
Experiment  also  shows  that,  from  this 
plane  towards  the  top  and  bottom,  the 
amount  of  extension  and  compression  may, 
for  the  stresses  that  occur  in  ordinary  prac- 


tice,  be  considered  as  varying  directly  with 
the  distance  from  the  neutral  plane. 

The  extreme  top  and  bottom  fibres  suffer 
the  greatest  compression  and  extension, 
and  in  case  of  rupture,  the  rupture  begins 
with  them.  Some  question  exists  as  to  the 
exact  location  of  the  neutral  line  or  plane, 
but  for  slight  deflections  it  passes  through 
the  centre  of  gravity  of  the  cross  section 
of  the  beam,  and  it  is  very  probable  that  it 
never  deviates  from  this  position. 

In  discussing  transverse  stress,  the  as- 
sumptions based  upon  experiment  may  be 
stated  as  follows : 

1.  The  forces  on  the  fibres  are  directly 
as  the  amount  of  extension  or  compression 
they  produce;   ( lit  tensio  sic  vis,)  and  since 
the  extension  and  compression  increase  as 
the   distance  from   the  neutral   axis,  the 
forces  vary  in  the  same  proportion. 

2.  Within   elastic  limits  the   extension 
and  compression  at  equal  distances  from 


10 

the  neutral  axis  arc  equal,   and  the  foices 
producing  them  are  equal. 

3.  The  neutral  axis  passes  through  the 
centre  of  gravity  of  the  cross  section. 

RECTANGULAR    BEAMS. 

Let  us  now  discuss  the  relations  existing 
between  the  forces  in,  and  on,  transversely 
loaded  rectangular  beams,  the  load  being 
supposed  to  be  vertical  in  direction  and 
the  beam  horizontal. 

Case  I. 

Let  A  D  (Fig.  2)  be  a  beam  so  thin  that 
it  may  be  considered  as  composed  of  but 
one  layer  of  fibres  or  particles.  Let  it  be 
fastened  in  a  wall  at  AB,  and  be  loaded 
with  W  at  the  other  end.  Neglect  for 
the  time  the  weight  of  the  thin  beam  it- 
self, which  is  small.  Imagine  it  to  be  cut 
by  a  vertical  plane  E  F  at  any  point,  and 
let  us  see  under  what  forces  the  part  E  D 
is  held  in  equilibrium. 

The  only  external  force  on  E  D  is  W 
acting  at  D  downwards,  and  E  D  is  pre- 


11 


G<* 

d 


vented  from  falling  under  this  weight  by 
the  resistance  of  the  fibres  at  EF.  To 
analize  these  forces,  let  us  take  O,  as  an 
origin  of  co-ordinates,  and  OlOf  as  the 
axis  of  x,  and  Ol  E  as  the  axis  of  y,  and 
as  the  forces  are  all  in  one  plane,  find 
their  components  along  these  axes.  The 
internal  forces,  or  resistance  of  the  fibres 
at  E  F,  are : 


12 

1.  The  horizontal  forces  which  are  ten- 
sile above  and  compressive  below,  and 
which  increase  from  zero  at  Oj  just  in  pro- 
portion as  we  go  from  that  point  towards 
the  upper  or  lower  edge  of  the  beam.  (Fig. 
3.) 


w 

(FiG.  3.) 

2.  The  vertical  force.  This  is  called  the 
shearing  force  or  transverse  shearing  force. 
It  resists  the  tendency  of  the  part  of  the 
beam  E  D  to  slide  down  on  the  surface  E 
F.  The  existence  of  this  force  may  be 
realized  if  we  conceive  the  beam  to  be 
divided  into  two  parts  by  the  vertical  plane 
E  F,  and  those  parts  to  be  united  by  some 
very  elastic  substance,  as  india-rubber. 


13 

Then  the  beam  would  take  the  form  shown 
in  Fig.  4,  the  part  F  C  sliding  down  on  the 


I 


other.  The  force  in  the  beam  that  resists 
this  sliding  is  represented  in  Fig.  3,  by  the 
vertical  arrow  at  E.  Let  it  be  called  T.  In 
Fig.  3  are  represented  all  the  forces  we 
have  to  deal  with.  Since  this  system  of 


14 


forces  is  balanced,  the  following  equations 
must  be  fulfilled  : 

2X=0     2Y=Q     J2M=0          (1) 

That  is :  the  sum  of  all  the  horizontal  forces 
(^X),  and  the  sum  of  all  the  vertical 
forces  (2  Y)  must  each  be  equal  to  zero, 
and  the  sum  of  the  moments  about  any 
point  as  Ol  must  also  equal  zero. 

The  only  horizontal  forces  in  the  system 
are  the  two  triangular  groups  of  forces  E  O, 
H  and  F.  Oj  L,  representing  the  sum  of 
the  tensile  and  compressive  stresses  on  the 
fibres.  As  the  group  EO,  II  acts  in  a  di- 
rection opposite  to  that  of  the  group  FO, 
L,  and  as  the  algebraic  sum  of  the  two 
groups  is  zero  (^X=z:0),  the  groups  must 
be  equal  to  each  other.  This  is  indicated 
in  the  figure  by  the  equality  of  the  triangles 
EOjHandFC^L. 

The  vertical  forces  are  W  and  the 
shearing  force  T  at  E  F,  and  since 

J2Y=T— W'  =  0.          We  have 

T=W  (2) 


15 

Next  obtain  the  moments  of  all  the 
forces  about  O1  and  place  the  sum  of  these 
moments  =  zero.  Replace  the  tensile  and 
compressivc  forces  by  their  resultants.  The 
resultant  or  sum  of  all  the  tensile  forces 
represented  by  the  triangle  E  Ot  II  (Fig.  3) 
may  evidently  be  represented  by  the  area 
of  the  triangle  of  which  the  base  EOjis 
the  distance  over  which  the  forces  are  dis- 
tributed, and  the  altitude  E  II  is  the  stress 
in  the  outside  fibre.  Let 

S  =  this  stress  —  E  H 
and  d  =  E  F  =  depth  of  beam 

Then  area  E  O:  H^=i  S  c?=N=   resultant 
of  tensile  forces.     Similarly 


Area  F  O,  L  —^Sd—  T$'=  resultant  of 
compressive  forces, 

These  resultants  will  pass  through  the 
centres  of  gravity  of  the  triangles  E  Ol  H 
and  Ol  FL,  since  the  little  forces  of  which 
they  are  composed  are  represented  by  these 
triangles.  Hence  the  direction  of  JVwill 
intersect  O14E  at  a  point  G  (Fig.  5),  whose 


1G 


M 


c 

distance    from    O1  is  —  2/3  E  O1=f-.  -  =— . 

\$   £       o 

This  is  the  lever  arm  of  JV  about  Ox.     That 

of  N'  is  G'  Oj  also  =  —     Hence  the  sum  of 

3* 

the  moments  of  these  two  forces  about  Ol 
(since  they  both  tend  to  produce  left- 
handed  rotation)  is 


17 

The  force  T  since  its  direction  passes 
through  Oj  has  no  lever-arm,  and  hence 
its  moment  is  zero. 

If  the  distance  from  O'  to  Ol  be  called 
x,  the  moment  of  the  weight  W  is  —  -f-W 
x  (since  it  tends  to  produce  right-handed 
rotation), 


;—  -Srff  =  ^M  =  0 

6 


(3) 


So  far  we  have  considered  a  beam  whose 
breadth  is  that  of  only  one  row  of  fibres, 
but  a  beam  of  any  breadth  may  be  made 
up  of  a  number  of  such  slices  placed  side 
by  side,  and  if  b—  the  number  of  slices,  or 
breadth  of  the  beam,  arid  W=the  weight 
hung  at  the  end  of  it,  then  eq.  (3)  becomes 

Wx^^Sbd2  (4) 

This  discussion  is  general  and  will  apply 
to  any  section  as  well  as  to  E  F.  S  and  x 
are  the  variables  in  eq.  (4),  and  these 
quantities  will  have  different  values  at 
the  different  sections,  which  values  in- 


18 

crease  as  we  go  towards  AB  (Fig.  2),  but 
the  form  of  the  equation  will  evidently  be 
unchanged.  If  A  C  (Fig.  2)  be  =  /,  we 
have  for  the  section  at  A  B  (Fig.  2) 


-Sbd*  (5) 

6 


AB  is  the  section  of  greatest  stress,  and 
the  beam  if  overloaded  will  break  there. 

The   quantity    —  S  b  d2  is  called  the  mo- 

ment of  resistance  of  the  fibres,  or  moment 
of  the  internal  forces,  and  is  often  written 
Jl/for  brevity.  Wx  is  called  the  moment 
of  the  weight,  or  moment  of  the  external 
forces. 

Let  the  maximum  value  of  —  S  b  d9  (eq.  5) 
be  called  M0. 

We  may  illustrate  geometrically  the 
variation  of  the  moments  M  —  TP#,  and 
consequently  of  the  stresses  produced  on 
the  outside  fibres  from  A  to  C. 

In  Fig.  6  let  A  0  be  the  beam.  Take 
a  line  on  some  scale  to  represent  the  value 


19 


FIG.  6. 


of  M0  =W£,  and  lay  it  off  from  A  per- 
pendicular to  A  C.  Let  AL  be  this  line. 
Draw  L  C.  Then  the  dotted  perpendiculars 
in  the  triangle  LAC  will  represent  the 
moments  of  resistance  in  the  beam  at  the 
several  points  at  which  they  are  drawn. 

From  eq.  (2)  it  is  seen  that  the  shearing 
force  is  constant  at  every  section  of  the 
beam.  This  force  we  may  assume  with 
sufficient  accuracy,  for  our  present  purpose, 
to  be  uniformly  distributed  over  the  cross 
section  of  the  beam  on  which  it  acts. 
Hence  if  A  =  area  of  cross  section  and, 


20 
shearing  force  on  a  unit  of  the  surface, 

T  =  W  =  A*  (6) 

Lay  off  A  C  (Fig.  7)  =  I  and  CP  =  W. 


1 

I 

I 

FIG.  7. 

Then  the  rectangle  AP  represents  geo- 
metrically the  shearing  stress  at  every 
point  of  the  beam. 

Corollary.  When  several  weights  as 
W  WjW  Fig.  8)  are  suspended  from  the 
beam  at  different  points,  the  moment  of 
resistance  at  any  point  is  equal  to  the  joint 
moments  of  the  weights  at  that  point. 
Thus,  calling  distances  measured  from  C, 
G,  and  E  towards  A,  x,  x\  and  x2  respect- 


21 


f — a^ 


o> 

I 


ively,  we  have  for  the  equation  of  moments 
for  points  between  C  and  G 

Wx  =  ~  S  b  d'2        Between  G  and  E 

D 

Waj+W.a;,  =-Sbd*   While  at  K, 
6 

for  instance,  it  is 


•22 

i^i   4  ,^2  —  ~ 

The  shearing  force  at  K  is 

T— W-hWj-f-Wa  (8) 

Geometrically.     Let  A  C  (Fig.  8)  =  I A  G 


=  1,  and  AE  =  J2.  Lay  off 
L  H  =  W4 1,  and  A I  =  W2 12.  Draw  the 
triangles  as  in  Fig.  8.  Then  N  P  =  total 
moment  at  K,  for  instance. 

The  shearing  force  is  represented  by  the 
rectangles  A  P,  N  O,  and  S  T  (Fig.  9),  and 
at  any  point  in  the  beam  is  equal  to  the 
sum  of  the  weights  between  that  point 
andC. 

EXAMPLES. 

(1.)  Suppose  the  safe  stress  per  square 
inch  to  be  1,000  Ibs.  (=S),  and  Z=10  ft., 
b  —  3  inches,  and  d  =  12  inches,  what 
weight  will  the  beam  support  ? 

(2.)  Suppos3  W=  8/4  ton,  1=12  ft.,  b=2 
inches,  what  must  be  the  depth  (d)  of  a 
rectangular  cast  iron  beam,  so  that  S  shall 
not  exceed  4  tons  ? 


23 


p          A* 


I 


Let  the  beam  be  as  in  the  last  case,  but 
with  the  load  distributed  uniformly  over  it 
(Fig  10).  Let  w  —  weight  on  a  unit  of 
length,  W  =  total  weight  on  A  C,  1=  A  C 
=  length,  d=  A  B  =:  depth,  x=  E  C  as  be- 
fore. Then  the  forces  to  be  considered  are 
represented  in  Fig.  11,  the  little  arrows 


24 

along  E  C  representing  the  weight  distrib- 
uted along  the  beam. 


I 


1 

0 


B 


FIG.  10. 

Replace  the  weights  along  E  C  by  their 
resultant,  which  is  =  w  x,  and  which  should 
be  applied  at  the  middle  point  of  E  C,  since 
the  little  weights  on  the  beam  are  uni- 
formly distributed.  Then  putting  the 
resultants  N  and  N'  in  place  of  the  tensile 
and  compressive  force,  and  proceeding  as 
before,  we  have 

T— wx=0 

T—w  x  (9) 


25 


Also  wx.-—~ 


~M     (10) 
At  AB  (Fig.  10)  these  equations  become 

<") 


1 

) 


By  comparing  the  last  equation  with  eq. 
(5),  we  see  that  if  the  weight  and  beam  be 
the  same  the  stress  on  the  fibres  in  this  last 
case  is  only  one-half  what  it  was  in  the 
former,  or,  what  amounts  to  the  same,  the 
beam  will  bear  twice  as  much  distributed 
over  it,  as  it  will  when  the  weight  is  con- 
centrated at  the  extremity. 

From  eq.  (9)  we  see  that  the  shearing 
force  is  not  constant  as  in  the  last  case,  but 
varies  as  x.  It  is  greatest  at  A. 

Geometrically.  The  equation  M  =  %  w  x2 
corresponds  with  that  of  a  parabola  with 
vertex  at  C  and  axis  vertical.  Lay  off  A  L 
(Fig.  12)  =i?of,  and  through  L  and  C 


o 

T 


LLLLUUUiiU, 


-is 


FIG.  11. 


27 


draw  a  parabola.  The  ordinates  of  this 
parabola  (dotted  in  the  figure)  will  repre- 
sent the  moments  at  the  several  points. 


The  equation  T  =.w  x  is  represented  by 
the  triangle  A  P  C  (Fig.  13),  which  there- 
fore gives  the  shearing  stress  at  every  point 
of  the  beam  when  A  P  is  taken  -=.wl. 

Corollary  1.  When  the  load  is  distrib- 
uted over  only  a  part  of  the  beam  as  in 
Fig.  (14),  let  R  C=m=the  loaded  part, 
and  take  the  other  letters  as  before.  Then 
the  equations  for  any  section  in  the  loaded 
part  are  evidently  the  same  as  those  just 
obtained,  viz. : 


28 


FIG.  13. 


O 


3 
O 


ffl 


29 

(12) 


At  R  V 

And  T=wx 

But  at  any  section  E  F  between  A  and 
R  the  moment  of  the  load  is  =  wm 
(x — Vs  m)>  the  latter  factor  being  the  dis- 
tance from  the  centre  of  gravity  of  the 
load  to  the  section  E  F.  The  moment  of 
resistance  having  the  same  form  as  before, 
we  have  for  the  equation  of  moments  for 
any  section  in  R  A 


wm  (x—l/2m)      -Sbd*  =M     (13) 

At  A  this  becomes 

w  m  (I — \  m)=M0  the  greatest  moment. 

The  shearing  force  at  E  F  being  equal 
in  amount  and  opposite  in  direction  to  the 
whole  load  between  E  F  and  C  will  be 

T=w  m  (14) 

Geometrically.  For  the  moments:  lay 
off  A  C  =  I  and  C  R  —m  (Fig.  15).  At  R 

erect  PR—    0    and  at  A  make  AL=:M0. 

/v 


30 


Through  C  and  D  draw  a  parabola  as  in 
the  last  case,  and  (since  eq.  [13]  is  of  the 
first  degree)  through  D  and  L  draw  a 
straight  line.  Then  from  C  to  R  the  ordi- 
nates  of  the  parabola  represent  the  mo- 
ments, and  from  R  to  A  they  are  represented 
by  the  ordinates  of  the  trapezoid  R  L.  For 
the  shearing  stress  (Fig.  16),  lay  off  from 


31 


<q  fc, 

R,  R  N=w  wi.  Draw  C  N  and  N  P.  Then 
the  triangle  CRN  (corresponding  to  the 
equation  T=wx)  gives  the  shearing  force 
at  each  point  in  C  R  while  the  rectangle  R 
P  (corresponding  to  the  equation  T—w  m) 
gives  the  force  in  the  remaining  segment 
of  the  beam. 

Corollary  2.     When  there  is  a  load  W 
at  the  extremity  C,  in  addition  to  the  load 


uniformly  distributed  over  the  beam  (Fig. 
17),  we  have  a  combination  of  Cases  I.  and 


B 


E 


a 


E 


FIG.  17. 


II.  and  the  moment  of  the  external  forces 
at  any  section,  E  F,  is 


Hence  the  equation  of  moments  is 


-f-W  x         (15) 


M=-S5<f= 
6 

This  is  greatest  at  A,  or 

M  =-S  b  d  =i  w  l*+  W  I 
6 

The  shearing  force 
At  A 


(16) 

(17) 
(18 


33 


Geometrically.  The  simplest  way  of 
representing  the  moments  is  to  construct 
those  due  to  each  kind  of  weight,  and 

qn  /v»* 

then  combine  them.     Thus,  let  M'  =  —— 

and  M"  =W  x.  Construct  M'  as  in  Case 
II.,  it  being  represented  by  a  parabola 
with  vertex  at  C  and  axis  vertical,  and  M" 
as  in  Case  7".,  it  being  represented  by  a 
triangle  (placed  under  A  C  for  conven- 
ience). Since  M  =  M'  -f-  M"  from  eq. 
(15)  we  have  the  total  moment  at  any  point 
E  (Fig.  18)  represented  by  the  sum  of  the 
ordinates  of  these  two  figures=N  P  Fig. 
(18.) 

The  moments  may  also  be  represented 
by  the  parabola  corresponding  to  eq.  (15) 
as  in  Fig.  (19.)  This  parabola  has  its  ver- 
tex at  C'  and  not  at  C.  Of  course,  only 
that  part  of  the  curve  between  C  and  A  is 
applicable  to  our  purpose. 

The  shearing  force  is  represented  by 
adding  the  triangle  N  PP'  (Fig.  20)  repre- 
senting the  variable  part  w  x  of  T  to  the 
rectangle  C  P  which  represents  the  constant 
part  W  of  T. 


34 


EXAMPLE. 

Discuss  the  forces  when  the  load  is  distrib- 
uted as  shown  in  Fig.  (21)  assuming 
various  values  for  LN  and  N  C  as  well  as 
for  w  and  W. 

Case  IIL 
Let  the  beam  whose  length  is  I  rest  upon 


35 


supports  at  B  and  D  (Fig.  22)  and  let  it  be 
loaded  at  some  point  G  with  a  single 
weight  W.  Let  m  and  n  be  the  segments 
into  which  the  beam  is  divided  at  the  point 
G  of  the  application  of  the  weight. 
.First  find  the  proportions  of  the  weight 


36 


2 
fa 


supported  at  B  and  D,  or  in  other  words, 
the  reactions  of  the  supports.  By  the  prin- 
ciple of  the  lever  the  respective  portions  of 
-  the  weight  supported  at  B  and  D  are  in- 
versely proportional  to  the  distances  of  these 
points  from  G.  Thus,  let  W'  — reaction  at 
D  and  W"  =  reaction  at  B  and  then 


W  :  W":  :m:n 
\y"-f  W :  W' :  :  m+n :  m 


o  tv 

ot 


g  g 


a  a  o 


FIG.  21. 


1 

I 


FlG. 


38 
But  W'+W"=Wand  m  +  n=l 


And  so        W"=4w 

I 

Now  apply  the  conditions  of  equilibrium 
to  any  part  A  E  of  the  beam  counting  from 
.A.     (Fig.  23), 


H 


2 
fe 


S 


39 


1st.  Between  A  and  G.  ^'X~0  merely 
indicates  the  equality  of  X  and  1ST',  as  these 
are  the  only  horizontal  forces.  2  Y~0 
shows  that  the  shearing  force  at  the  section 
E  F  is  downwards 

and=W"  .  •  .  T=  W"  =  ^W  (20) 

2  M=  0.     The  joint  moment  of  N  and 

N'  is  as  before=-S  b  d\     That   of   T  is 
6 

zero.  The  only  other  force  acting  on  A 
E  is  W",  the  reaction  of  the  abutment. 
Let  O  Ol  —  x.  Then  the  moment  of  W" 
is 

W"X=      .Wa? 


11  1 

Hence      j.Wx—  ± 

.•~Wx=-Sbd2  =  M  (21) 

i  6 

2d.  Between  G  and  C  (Fig.  24).  Here, 
between  A  and  E  are  the  two  external 
forces  W"  at  B  and  W  at  G.  Hence  the 
shearing  force  at  E  F  is  upwards  and 


40 


P         S3 


Sz; 


T= W— W=  — W'  =  — yW.  (22) 

For  >'  M  =rr  0  we  have 

-  -  S  b  c72+  W  'x—W  (x—m)  =  0 
.  • .  y Wx  —W  (x—m)  = l-  S  b  d2—  M       (23) 

fr  U 


41 


The  greatest  value  of  eq.  (21)  is  at  G, 
where*  it  becomes  identical  with  eq.  (23) 
for  the  same  point.  This  value  is 


At  A  and  C  the  moments  are  zero. 


Geometrically.   From  eq.  (21),  which  is  of 
the  first  degree,  it  is  seen  that  the  moments 


42 

vary  in  A  G  as  they  did  in  Case  I.  Hence 
they  may  be  represented  by  the  ordinates 
of  the  triangle  A  L  G.  (Fig.  25). 

Eq.  (23)  is  also  that  of  a  straight  line 
cutting  the  axis  of  X  at  C.  Hence  the 
moments  in  G  C  are  represented  by  the 
triangle  G  L  C  (Fig.  25).  The  shearing 


I 


43 


force  in  A  G  is  represented  by  the  rectangle 
A  P  and  that  in  G  C  by  the  rectangle  CP'. 
(Fig.  26.) 

Note.  That  the  maximum  moment  (at  G) 
corresponds  to  the  point  where  the  shear- 
ing force  passes  through  zero. 

Corollary  1.  When  the  weight  W  is  at 
the  middle  of  the  beam,  we  have 


Then  eq.  (21)  becomes 


V, 

and  (23)  is 


«-x,=\ 


(25) 


At  the  centre  M0=  Y2W.  V,  I  =  l/,W  I  (26) 
The  triangles  A  L  G  and  G  L  C  (Fig.  27) 

represent  the  moments  in  this  case.     G  L 

having  been  laid  off  ==.  */4  W  L 

The  shearing  force  throughout  the  beam 

is  then  T=l/2W  (27)  as  is  shown  in  the 

rectangles  (Fig.  28). 

Comparing  the  value  of  M0  given  in  eq. 


44 


(5)  Case  I.  with  that  of  M0  in  eq.  (26)  we 
see  that  the  load  and  the  length  being  the 
same,  a  beam  will  bear  four  times  as  much 
with  both  ends  supported,  and  the  load 
placed  in  the  middle  as  it  will  do  with  one 
end  fixed  and  the  other  loaded. 

Corollary  2.     When   there   are   several 
weights,  as  in  Fig  (29),  the  moment  of  the 


external  forces  at  any  section  is  that  due 
to  the  action  of  all  the  weights.  Let  the 
segments  into  which  the  weights  W,  W, 
and  W8  divide  the  beam  be  m  and  n  for 
W,  ml  and  nl  for  Wl  and  m2  and  n^  for 

wf. 

Let  Ra  —  reaction  of  abutment  at  B  and 
R2=  reaction  of  abutment  at  D  and 
£= length  and  let  x  be  counted  from 
A  as  before. 


46 


>.S 

^  ^ 

v 

s 

., 

a 

1 

t-/t% 

%h 

- 

X 

;  y| 
/^% 

, 

a 

' 

<J 
i__/^ifc 

y 

s 

> 

-     V 

tt 

£ 

,  > 

9 

<i      PQ 

|f^^W 

The  reaction  of  the  abutment  at  B  due 
to  these  weights  is 


???__..      m  ^      ,  m9^ 
So  at  D  R2=z—  W+  7  -'Wj+^Wg 

It  i 


t 


For   any  section   between  A  and  G,  Rj 


'47 

is  the  only  external  force  and  hence  the 
equation  of  moments  is 

R^ifi  b  cf=M  (29) 

The  shearing  force  is        T=R,l  (30) 

For  every  section  between  G  and  G' 
the  weight  W  is  to  be  taken  into  consider- 
ation and  the  equations  are  : 

R^—  W(x—  m)=^S  b  <F=M        (31) 

TznR,—  W  (32) 

For  any  section  between  G'  and  G") 
we  have  : 

(33) 


T=R1_W—  W, 

For  any  section  between  G"  and  C 
Rix—W(x—m)—W1(x—m})—W2  \ 

(x—m^)=^Sbd2  =  'M.  I    (34) 

T=Rl—W—  W,—  W,  ) 

The  location  of  the  greatest  moment  is 
most  readily  determined  by  geometrical 
construction. 


48 


Geometrically.     The  moments  are  rep- 
resented in  Fig.  (30)  by  constructing  sep- 


o 

CO 


arately  those  due  to  each  weight  and  then 
combining  them.  Thus,  the  moments  pro- 
duced at  every  point  in  the  beam  by  the 
weight  W  are  represented  by  the  triangle 


>    49 

A  L  C  (Fig.  30),  in  which  G  L  equals  the 
greatest  moment  due  to  W=—  W  m.  Sim- 
ilarly A  L'  C  represents  those  due  to  W,, 

n 
G'  I/  being  =— 1W1m1    and  A  L"  C  gives 

?i 
those  due  to  W2,  G"  L"  being=i-2  W2  m2. 

Now,  if  at  every  point  we  add  together 
the  ordinates  of  these  three  triangles  for 
that  point,  and  lay  them  off  above  A  C  we 
shall  get  a  polygon  A  H  H'  H"  C,  which 
represents  eqs.  (29)  (31)  (33)  and  (34)  and 
gives  the  total  moment  at  any  section.  The 
greatest  ordinate  of  this  polygon  will,  of 
course,  show  the  location  of  the  maximum 
moment.  This  will  be  at  G  or  G'  or  G"> 
according  to  the  relative  amounts  and  pos- 
itions of  the  weights  W,  W,,  and  W2.  In 
the  Fig.  it  is  at  G'.  Hence  from  eq.  (31) 

R  ml—W(ml—m)=-.  S  b  d2  =  M0    (35) 
The  shearing  force  may  be  represented 


50 


3 

£ 


..*.... 


~" 


as  in  Fig.  (31).  It  is  greatest  in  that  one 
of  the  two  end  segments  which  corresponds 
to  the  greater  of  the  two  quantities  Ra  and 


Note.  That  in  this  case  the  simplest  way 
of  finding  the  point  of  maximum  moment 


is  to  construct  the  figure  representing  the 
shearing  force,  and  the  point  when  the 
shearing  force  passes  through  zero,  (G' 
Fig.  31)  is  the  point  sought. 


EXAMPLES. 

1.  Let  fc20  ft.,  m  —  6  ft.,  ^  =  10  ft.,  wf= 
15  ft.,  W=l  ton,  W,  =  2  tons,  W2=3  tons. 

Find  the  maximum  moment. 

2.  Find  the  size  of  a  rectangular  wooden 

beam  where 

/=15  ft.,  m=3  ft.,  m1=.  6  ft.,  wi2  =  14  ft. 
W=l  ton,  W,=i  ton,  Wk=2  tons 
8=1,000  Ibs.  and  d—^  b. 

Case  IV. 

Let  the  load  be  equally  distributed  over 
the  beam  (Fig.  32).  In  this  case  the  reac- 
tion of  each  abutment =%  the  load,  or 

R,=^=K,  (36) 

rj 

Take  any  section  E  F  whose  distance 
from  A=&.  Then  the  external  forces 


CC 

6 


acting  between  A  and  E  F  are,  R1  and  the 
resultant  of  all  the  little  weights  from  A 
to  E  (—w  x).  This  last  force  acts  at  its 
centre  of  gravity  (Fig.  33),  which  is  half 
way  from  A  to  E.  Its  lever  arm  is  there- 

oc 

fore  =  —     Hence  the  equation  of  moments 
& 

will  be 


53 


CO 
CO 


or 


(37) 


This  is  a  maximum  at  the  centre  where 
M0=  »/.  to  P  •  (38) 

The  shearing  force  at  E  F  is 


T=- wx  (39) 

This  is  greatest  at  the  abutments  where  x 
—  I,  or  0 


.   T  _ 
°~  2 


(40) 

At  the  centre         T=0 

Geometrically.     The  values  of  M  in  eq. 
(37)   may   be   represented  by   a  parabola 


^    55 

with  vertex  at  L,  the  ordinates  G  L 
(Fig.  34)  being  taken  =  M0.  The  shear- 
ing force  is  represented  by  the  two  triangles 
A  P  G  and  G  Q  C  (Fig.  35). 


- 


The  maximum  moment  exists  at  the  point 
(G)  when  the  shearing  force  is  zero. 


56 


Corollary.  When  the  uniform  load  ex- 
tends only  over  a  certain  distance  from  one 
of  the  supports,  as  in  (Fig.  36),  let  A  T>= 


CO 

CO 

A 


loaded  segment—  m.     The  reactions  of  the 
abutments  are: 

At  A,  R^w  ml  -  Y~  )  ) 

V      l     ' 


At  C,  l&i= 


(41) 


Then  for  any  section  in  A  D  the  moments 
of  the  external  forces  will  be  as  in  the  case 
just  discussed. 


l 

\ 


w 


And  the  equation  of  moments  in  A  D  will 
be: 


w  m 


tl—%  m\        wx*     1 

(          -  —  x—  —=-  S  o  a  =M      (42) 
\      I      /          2        b 

For  any  section  in  D  C  the  whole  load 
(w  m)  is  to  be  considered  as  acting  through 
its  centre  of  gravity  (G),  and  the  equation 
of  moments  is: 


fl  -        m\  ,  .  1     C1     7        72          T»/T 

l  —  —  \x  —  wm(x>  —  %m)=—  S  o  a  =M 


Reducing 

'^f(l—x)  =I  S  b  c72==M  (43) 

/v    6  O 

The  shearing  force  in  A  D  is 

/l=$  m\  ^ 

L  —  w  m.l  -  -  —  J  —  w  x  I 

(44) 
In  D  C,  T=w  m. 


58 
T  is  a  maximum  at  A,  or 


Geometrically.     Eq.  (42)  corresponds  to 
the  parabola  A  L  K  (Fig.  37),  which  cuts 


A  C  at  A  and  K  (whose  distance  from  A= 
— y-  [I — i  m]  )  and  whose  axis  is  vertical 


59 


Eq.  (43)  corresponds  to  the  straight  line  H 
C.  We  can  only  use  the  part  A  L  H  of 
the  parabola,  the  moments  in  D  C  being 
represented  by  the  triangle  D  H  C.  The 
maximum  moment  is  at  N  corresponding 
to  the  vortex  L  of  the  parabola.  The  value 
of  this  moment  is: 


(45) 


which  is  obtained  from  eq.  (42)  by  substi- 
tuting for  x  the  value  A  N"  (=4  A  K)= 

tn 

*—(l  —  4  m).     This  M0  is  always   less  than 

the  maximum  moment  that  exists  when  the 
load  extends  all  over  the  beam  as  will  ap- 
pear by  making  m  to  vary  in  eq.  (45)  and 
applying  the  tests  for  a  maximum  to  it. 

The  shearing  stress  for  the  loaded  seg- 
ment is  represented  by  the  triangles  A  P  K 
and  K  P'  D  (Fig.  38),  and  for  the  other 
Segment  by  the  rectangle  D  H.  The  point 
K,  where  the  stress  is  zero,  is  found  by 
making  in  eq.  (44) 


60 

rp  il—\m\ 

I  =10  m  f  —  —  * — w  x=0 

and  finding  the  value  of  x. 


This  point  corresponds  to  the  maximum 
moment.  It  may  also  be  found  graphi- 
cally by  constructing  Fig.  38,  and,  as  be- 
fore, affords  the  easiest  method  of  deter- 


mining  the  point  of  the  beam  where  the 
maximum  moment  exists  and  where  conse- 
quently there  is  greatest  danger  of  rupture. 


EXAMPLES. 


1.  Let  1=20  ft.  -20=500  Ibs.  per  ft.  m= 
15  ft.  and  let  there  be  a  weight  in  addition. 


G 
A 

)    @  Q    Q  ©  Q   © 

c 

1 

i 

f 

! 

TV 

FIG.  39. 


W=5  tons  at  a  point  18  ft.  distant  from 
A.     Required  the  maximum  moment. 


AT 


FIG.  40. 


62 


2.  Let  one-half  of  the  above  beam  be 
loaded  with  a  uniform  load,  w=~L  ton  per 
foot,  and  the  other  half  with  a  uniform 
load  of  w  =i  ton  per  foot.  Required 
the  moments. 


Case  V. 

A  single  moving  load.  When  a  single 
moving  load  passes  over  a  beam,  as  in 
(Fig.  41),  the  maximum  moment  at  each 
instant  (as  appears  from  Case  III.)  takes 
Affect  at  the  section  just  under  the  weight. 
To  determine  the  law  of  variation  of  these 
maxima  as  the  weight  travels  over  the 
beam:  Let  rc=the  distance  at  any  instant 
from  A,  and  then  the  reaction  of  A  at  that 
instant  (=  the  part  of  W  transmitted  to  it) 

I x 

—  W  —  —     Multiplying  this  by  the  lever 

V 

arm  x  we  have  for  the  moment  under  the 

W   7*  1 

weight :  — —  (l-x)  =  ~  S  b  da=M  (46) 

This  is  a  maximum  at  the  centre,  where 


Eq.  (46)  corresponds  to  a  parabola  (Fig  41) 
with  vertex  at  L,  the  ordinate  G  L  being= 
V4  W  L  The  shearing  force  for  each  seg- 
ment into  which  W  (Fig.  42)  at  any  instant 
divides  the  beam  is  equal  to  the  reaction  of 


61 

the  abutment  corresponding  to  that  se# 
ment.     Thus,  if  W  is  at  a  distance  x  from 


A  the  reaction  of  A  is  =W~ ^Land  of  C 


Tx 

r 


If  W  has  the  position  marked  2  in  (Fig. 
42),  then    the  shearing  stress  in  the  left 


I 


^       <9      P4        fe 


segment  is  shown  by  the  rectangle 
W  and  in  the  right  segment  by  the  rec- 
tangle WN"  N"'  C.  The  diagram  shows 
in  a  similar  way  the  stress  at  other  points. 
If  the  third  position  of  W  in  the  figure  is 
at  the  centre  of  the  beam  then  evidently 
the  greatest  shearing  stress  to  be  provided 
for  in  the  left  half  of  the  beam  will  be 
represented  by  the  locus  of  the  points  like 
L,  N',  P',  and  for  the  right  half  it  will  be 
the  locus  of  the  points  P",  Q",  L',  etc. 

These  loci  are  given  by  the  equations  : 

W 

T=— «— *)= eq.  of  L  C  ) 

^  (48) 

T=— x         =eq.  ofALM 

L  *  ' 

Turn  the  triangle  A  L'  C  down  for  con- 
venience, as  in  (Fig.  43),  and  then  the 
shearing  stress  to  be  provided  for  is  given 
b.y  the  figure  A  L  P  I/  C.  In  this  figure 

AL  =  CL'  =  W  and  D  P==^. 


Case   VI. 

A  distributed  moving  load.  When  a 
moving  load  gradually  covers  a  beam, 
(Fig.  44),  moving  on  from  one  end  as  a  long 
train  of  cars,  the  maximum  moments  pro- 
duced is  that  due  to  the  load  when  it  covers 
the  entire  length  of  the  beam,  and  conse- 


67 


o 

£ 


queritly  this  case  is  provided  for  in   Case 
IV. 

But  with  the  shearing  stress  it  is  dif- 
ferent. Here,  as  in  Case  V.,  we  need  the 
locus  of  the  greatest  shearing  stresses  that 
be  brought  upon  the  beam.  This  maximum 
at  any  section  D  occurs  when  the  longer 
segment  into  which  D  divides  the  beam  is 


68 


loaded,  and  the  other  is  not.  In  that  case 
the  shearing  force  at  D  (=the  reaction  of 
the  abutment  C)  is 

T-—  (49) 

~   2  I 

This    equation    gives    the   parabola    A 
N  P'  (Fig.   45)  with   vertex  at  A,  where 


'    69 

nn     7 

CP'  =  -—  .     When  the   load    comes   from 

the  other  end  of  the  beam  we  get  the 
parabola  C  N  P.  Hence  the  figure  A  P 
N  P'  C  gives  the  maximum  shearing  stress 
to  be  provided  for. 

It  is  easy  to  see  that  the  shearing 
stresses  thus  obtained  are  greater  than 
those  which  exist  whep  the  load  covers 
the  entire  beam.  In  the  latter  case  the 
forces  are  represented  by  the  triangles 
A  P  G  and  G  P'  C  (Fig.  45),  the  shearing 
'  stress  at  D  being  given  by  eq.  (39J 


In   the  case  of  the  passing  load  we  have 
just  seen  that 

T     WX*     T»K 
~l 

/C     i 

The  value  of  D  H  is  always  less  than  that 

of  D  K  when  x>_-^  ;  for  if  2  x  be  a  certain 
« 

quantity,  then   the  product  of  the  halves 
of  that  quantity  (=x2)  is  greater  than  the 


product  of  any  other  two  parts  ('such 
as  I  and  [2  x—l\  )  into  which  it  can  be 
separated. 


That  is 


w  x2      w  I  (2  x—l 
TT>  2  I 


W  X 

TT>  * 


(50) 


In  the  expressions  of  the  moment  of 
resistance  M=—  S  b  d~  the  quantity  de- 

noted by  S  (  =  the  stress  on  the  outside 
fibres)  varies  directly  as  M.  Hence,  all 
the  geometrical  illustrations  we  have 
given  of  the  moments  may  apply  equally 
well  to  the  values  of  S.  The  maximum 
moments  give  the  maximum  stress  on 
the  fibres,  and  indicate  the  points  of  rup- 
ture when  the  beam  is  loaded  with  its 
breaking  weight. 

ULTIMATE    VALUES    OF     S. 

If  beams  are  loaded  transversely  until 
fracture  takes  place,  the  value  of  S  or  the 
stress  on  the  outside  fibre  which  exists  at 


71 

the  moment  of  fracture,  gives  us  a  value  for 
the  tensile  or  compressive  strength  of 
the  material  according  to  the  manner  of  rup- 
ture. If  the  beam  yields  by  tearing,  S 
gives  us  the  tensile  strength,  if  by  crushing 
S  gives  the  compressive  strength.  We 
readily  obtain  the  value  of  S  answering 
to  the  ultimate  strength  from  any  of  the 
formulas  under  "  Transverse  Stress,"  by 
substituting  given  values  for  /,  #,  and  6?, 

and  the  actual  breaking  weight  for  W. 

- 

But  the  tensile  and  compressive  strengths 
of  materials  are  also  obtained  by  direct 
tension  and  compression,  the  force  being 
applied  in  the  direction  of  the  length 
of  the  .bars  until  rupture  takes  place. 

If  our  theory  were  perfect  the  values 
of  tensile  and  compressive  strength  thus 
deduced  would  agree  with  the  ultimate 
values  of  S  found  in  tranverse  stress;  but 
they  do  not. 

The  difference  is  very  wide  sometimes. 
Thus  in  cast  iron,  S  (in  this  case  it  repre- 
sents the  tensile  strength)  derived  from 


breaking  rectangular  beams  by  a  trans- 
verse load  is  nearly  20  tons  per  square 
inch,  while  the  tensile  strength  obtained 
directly  is  only  about  8  tons.  This  dis- 
crepancy has  been  accounted  for  in  two 
ways. 

1.  That  the  neutral  axis  moves  towards 
the  compressed   side,    and    that   therefore 
a  larger  portion  of  the  beam  is  subjected 
to  tension  than  the  formula  supposes. 

2.  That  the  neutral  axis  always  remain- 
ing at  the  centre  of  gravity  of  the  beam, 
the   additional    strength    is    due    to    the 
adhesion  of  the  fibres  which  is  developed 
by   the   unequal   lengthening,    and    short- 
ening of  them  as  we  go  from  the  neutral 
axis  towards  the  surfaces.     In  favor  of  this 
view   is   the  fact   that  we  know  such  ad- 
hesion to  be  an  element  of  strength;  for 
the    compression    or    extension    due   to  a 
given   force   is   not   so   great   in   a  trans- 
versely  loaded    beam   as   in    one    directly 
compressed  or  extended. 

The  action  of  this  adhesive  force  may  be 
illustrated  as  follows: 


FIG.  46. 

In  the  beam  A  B  (Fig.  46),  strained  by 
the  weight  W,  all  the  fibres  are  equally 
elongated,  and  they  only  resist  by  their 
direct  tenacity.  But  in  the  beam  A'C 
to  the  one-half  of  which  is  appended  the 
weight,  while  the  other  half,  E  0,  is  less 
strained  or  altogether  prevented  from 
extending,  evidently  W  will  have  to 
overcome  not  merely  the  tenacity  of  the 
fibres  in  A'  B'  but  the  adhesive  force  of 


74 

the  fibres  along  the  plane  E  F,  where  the 
two  parts  of  the  beam  join ;  for  this  force 
will  tend  to  prevent  the  stretching  of  the 
fibres  in  A'  13',  and  consequently  increases 
the  strength  of  A'  B'.  This  kind  of  force 
exists  between  every  two  layers  of  hori- 
zontal fibres  in  a  beam  under  transverse 
loading,  and  is  called  the  longitudinal 
shearing  stress.  It  is  neglected  in  the 
formulae  we  have  given. 

From  the  variation  between  the  ulti- 
mate values  of  S  (called  moduli  of  rupture) 
and  the  values  for  strength  obtained  by 
direct  tension  and  compression,  it  results 
that  the  values  should  be  determined  in 
both  ways,  and  that  the  values  gotten  by 
one  method  should  not  be  used  in 
calculations  involving  the  other  kind  of 
stress. 

BEAMS    OF    UNIFORM    STRENGTH. 

As  already  stated,  in  solid  rectangular 
beams,  S  has  different  values  for  the  va- 
rious points  in  the  length  of  the  beam. 
There  is  always  a  point  of  maximum 


«.   75 

stress  where  the  beam,  if  loaded  suffi- 
ciently, will  break.  Now  at  all  other 
points  there  is  an  excess  of  material 
which  is  useless  and  injurious  from  its 
own  weight.  To  secure  the  requisite 
strength  with  the  least  material  is  an  ob- 
ject usually  desirable,  and  this  can  be 
readily  accomplished  in  certain  materials 
(as  cast  iron),  by  giving  the  beam  such 
a  shape  as  will  make  S,  the  stress  on  the 
outside  fibre,  constant  throughout  its 
length.  In  wood  the  injury  resulting 
from  the  cross  cutting  of  the  fibres  fre- 
quently prevents  the  putting  of  the  theory 
into  practice. 

The  application  of  the  theory  of  uni- 
form strength  to  beams  of  rectangular 
cross  section  may  be  most  simply  explained 
by  taking  up  the  cases  we  have  discussed 
in  detail. 

In  Case  I.  from  eq.  (5)  the  maximum 
stress  in  the  outside  fibre  is, 

(51) 
This  stress  only  occurs  at  A,  where  the 


76 

beam  will  ultimately  break,  and  it  is 
evidently  possible  to  take  away  some 
of  the  material  between  that  point  and 


FIG.  47. 

C  without  diminishing  the  strength.  If 
this  be  so  done  that  at  every  point  be- 
tween A  and  C  there  shall  exist  on  the 
outside  fibre  a  stress  equal  to  that  at  A, 
the  beam  will  be  one  of  uniform  strength, 
and  we  shall  have  attained  the  greatest 
economy  of  material.  Let  us  suppose, 
the  use  we  have  for  the  beam  requires 
the  depth  to  be  uniform.  What  must  be 
its  plan  in  order  that  S  shall  be  constant 
in  value,  or  the  beam  be  as  liable  to 
break  at  any  other  point  as  at  A6. 

In  eq.  (4)  S=-^- ™- ,  if  we   assume  S  to 


77 

be  constant,  the  other  side  of  the  equation 
must  be  constant  also,  and  since  6  Wis 
constant,  and  we  have  made  d  constant 
by  assuming  the  depth  to  be  uniform,  the 
whole  expression  can  be  constant  only 
when  b  varies  as  x. 

byox  whence  b=c  x  (where  c  = 
some  constant  factor).  This  equation 
which  is  that  of  a  straight  line  shows  that 
the  breadth  must  vary  directly  as  the 
length.  Hence  the  plan  should  be  a  tri- 
angle with  vertex  at  C  (Fig.  48). 


FIG.  48. 

On  the  other  hand,  if  we  suppose  the 
breadth  to  be  uniform  and  wish  to  have 

6  Wx 

S  constant,    in   the   eq.  S=    v~7i~>  x 

u  d 

vary  as  d,  or2 

d2—  c  x 


78 


This  corresponds  to  a  parabola,  and  the 
beam,  if  the  top  be  straight  will  have  the 
elevation  shown  in  (Fig  49). 

Suppose  that  b  varies  as  d,  then  d  =n  b 
(?i  being  a  constant)  and 


S= 


6  W  x 


79 

To  render  S  constant  we  must  have,  58x> 
x  .  • .  b*=cx  and  d*=nB  ex. 

These  are  the  equations  of  a  cubic  para- 
bola. Hence  the  horizontal  section  (Fig. 


FIG.  50. 


50),   and   the   vertical   section    (Fig.   51),, 
should  be  curves  of  that  kind.     The  cross 


FIG.  51. 


80 

section  is  rectangular  as  in  (Fig.  52). 
In  Case  II.  we  have 


FIG.  52. 

Hence,  if  we  make  suppositions  similar 
to  those  above  we  shall  have,  when  the 
depth  is  uniform  (or  d=&  constant)  cc2 
varying  as  6,  or 

b=cx* 

Hence  the  plan  (Fig.  53)  should  consist 
r  of  parabolas  with  vertices  at  C.  If  b  feS 
"  constant  then 

d^—  ex2  or  d=<\/ ex. 
This  is  the  equation  of  a  straight  line,  and 


81 


FIG.  53. 
gives  for  the  elevation  the  triangle  (Fig.  54). 


FIG.  54. 

In  Case  III.  the  analysis  gives  results 
>  i  milar  to  those  in  Case  I. 

6  " 
Thus  from  equation  (21)  S= 


x 
-,  7a- 

0Cf  . 


-?  -  . 
i 

TT       6W  n  . 

Here  —  -  —  is  constant,  and  if  a  be 


constant  also,  b  must  vary  as  x. 
.  •  .  b  =  ex. 


This  gives  the  triangle  A  G  H  (Fig.  55.) 
We  obtain  similarly  the  triangle  G  C  H  for 
the  other  end  of  the  beam. 


If  the   breadth  be   constant   we   have, 


=  ex- 


which  gives  a  parabola  A  K  (Fig.  56)  with 
vertex  at  A.  So,  for  the  right  hand  end 
of  the  beam  the  proper  elevation  is  the 


83 


parabola  C  K  (Fig.  56).  The  elevation 
(Fig.  56)  assumes  that  the  top  of  the  beam 
needs  to  be  horizontal. 


In     Case    IV.,     equation     (37)    gives 

S= — 7— r — —  •'     Here,  if  d  be  constant  in 
b  d 

order  to  render  S  constant  we  have 
b  =  cx(l — x). 


84 


This  may  be  represented  by  parabolas 
with  vertices  at  G  and  H  (Fig.  57)  oppo- 
site the  middle  of  the  beam.  If  b  is  con- 
stant, then, 


FIG.  57. 

which  is  the  equation  of  an  ellipse,  and 
the  beam  (if  it  is  required  to  be  horizontal 
on  top)  may  be  made  as  in  (Fig.  58). 

In  these  cases  of  beams  of  uniform 
strength,  we  have  so  far  only  considered 
the  moments  of  the  weights  or  the  bending 
moments  as  they  are  called.  But  the  re- 
sults are  to  be  modified  by  the  transverse 
shearing  stress.  In  ordinary  rectangular 


B: 

FIG.  58. 


beams  this  shearing  stress  is  so  small 
compared  with  the  bending  moment,  that 
it  may  be  left  out  of  consideration.  But 
in  beams  of  uniform  strength  the  ends  must 
not  taper  to  a  point,  but  must  always  be 
left  large  enough  to  bear  the  shearing 
stress.  In  the  case  represented  in  (Fig.  57) 
the  beam  should  have,  near  the  ends,  the 
shape  shown  in  (Fig.  59). 


86 

DOUBLED  FLANGED  BEAMS. 

So  far  we  have  considered  beams  with 
rectangular  cross  sections,  and  beams  of 
uniform  strength  deduced  from  these.  We 
will  not  consider  beams  of  x  shape. 

It  is  evident  from  the  investigation  al- 
ready given  of  the  condition  of  stress,  in 
transversely  loaded  beams,  that  those  por- 
tions of  the  beam  nearest  the  centre  bear 
but  a  small  proportion  of  the  stress,  while 
the  contrary  is  the  case  with  the  outside 
fibres.  Hence  we  would  gain  strength  by 
moving  a  considerable  portion  of  that 
about  the  neutral  axis  and  placing  it  on 
the  top  and  bottom. 

The  first  form  in  which  the  idea  was 
applied  was  in  the  T  or  _[_  cast  iron  beam. 
The  fact  that  rectangular  cast-iron  beam 
always  broke  by  the  tearing  of  the  fibres 
on  the  side  subjected  to  tension,  suggested 
the  idea  of  reinforcing  that  side  of  the 
beam  with  a  flange.  The  results  of  this  is, 
that  the  neutral  axis  still  passing  through 
the  centre  of  gravity  of  the  cross  section, 


8? 

the  extreme  fibres  subjected  to  compression 
are  farther  off  than  those  subjected  to  ten- 
sion, and  consequently  are  strained  more 
nearly  to  their  full  strength  before  fracture. 
This  form  of  beam  gives  a  large  increase 
of  strength  for  the  same  amount  of  iron. 

It  was  still  plain  that  the  fibres  in  that 
part  of  the  web  about  the  neutral  axis  were 
but  little  strained  as  compared  with  the 
fibres  on  the  outside,  and  it  was  proposed 
to  leave  as  little  material  there  as  possible, 
and  to  place  the  mass  of  it  in  two  flanges 
( i  ),  one  above  and  the  other  below,  giv- 
ing to  these  flanges  sizes  inversely  propor- 
tional to  the  tensile  and  compressive 
strength  of  the  material.  The  question 
then  was,  how  much  of  the  material  should 
be  left  in  the  web,  for  plainly  all  could  net 
be  taken.  The  amount  to  be  left  is  deter- 
mined by  experiment.  If  the  web  is  left  too 
thin,  the  beam  will  twist  and  break  under 
the  shearing  force,  and  in  some  cases,  from 
the  want  of  stiffness  in  the  compressed 
flange. 

To  simplify  the  calculations,  the  web  is 
consid  jred  as  bearing  all  the  shearing  stress, 


and  no  other,  and  the  flanges  as  bearing 
all  the  extension  and  compression  due  to 
the  bending  moment  ;  and  these  parts 
should  be  proportioned  accordingly  with 
due  reference  to  the  practical  difficulties 
that  sometimes  occur.  The  ordinary  for- 
mulas for  the  strength  of  such  beams  are 
gotten  by  the  following  approximation  : 
We  first  neglect  the  compressive  and  tensile 
forces  of  the  web,  which  are  small  com- 
pared with  those  of  the  flanges,  and  con- 
sider it  as  bearing  only  the  shearing  stress- 
Then  as  the  depth  of  the  flanges  is 
generally  small  as  compared  with  the 
depth  of  the  beam,  we  consider  all  the 
fibres  in  each  flange  as  strained  alike,  and 
as  bearing  the  average  stress  that  is  brought 
on  that  flange.  (Fig.  60.) 

The  resultant  of  the  force  on  each  flange, 
then,  is  equal  to  the  stress  on  a  unit  of  sur- 
face (S )  multiplied  by  the  flange  area  (A)  : 
that  is =S  A. 

The  point  of  application  of  the  force  will 
be  at  the  middle  of  the  depth  of  theflangea 


A. 
B 


89 


FIG.  60. 
(at  O  and  O',  Fig.  61).     Fig.  61  shows  the 


FIG.  61. 

forces  we  have  to  deal  with  in  the  Case 
corresponding  to  Case  I.  under  rectangular 
beams. 


90 

Let  O'  O  =  <7. 

S'  =  stress  on  upper  flange  per  unit 

of  surface. 
S"=stress  on  lower  flange  per  unit 

of  surface. 

A" = area  of  lower  flange. 
A'  =area  of  upper  flange. 

Then  if  we  take  O  (Fig.  61)  as  a  centre 
of  moments  we  have : 

— N  d  — N'.O.+W  x  =  0  (ButN=S'  A') 
.  •  .  S'  A'  d=  W  x  (57) 

If  we  take  O'  as  the  centre  of  moments 

we  will  get 

S"  A"  d  =  W  x  (58) 

The  formula  for  shearing  force  is  iden- 
tical with  that  under  Case  I.  of  rectangular 
beams;  that  is: 

T=  W  x  (59) 

If  A'  =  A",  then  plainly  S'  =  S"  (from 
equations  57  and  58),  or,  the  forces  of 
tension  and  compression  are  equal  (as  in 
rectangular  beams);  but  if  A'  and  A"  are 
not  equal,  we  have: 


91 


That  is,  the  unit  stresses  in  the  flanges  are 
inversely  as  the  areas.  Now,  to  have  the 
material  distributed  between  the  flanges 
most  efficiently  for  strength  the  unit  stress 
should  be  in  proportion  to  the  ultimate 
strength  of  the  material  against  tension 
and  compression,  and  hence  the  areas  of 
the  cross  sections  of  the  flanges  should  be 
inversely  as  the  ultimate  strength. 

Thus,  if  A  D  (Fig.  62}  be  of  cast-iron, 
which  is  six  times  as  strong  against  com- 
pression as  against  tension,  the  unit  stress 
in  the  lower  flange  should  be  made  six 
times  as  great  as  in  the  upper,  and  to  effect 
this  the  area  of  the  lower  flange  should  be 
one-sixth  that  of  the  upper. 

The  Casexu  nder  x  beams  are  similar  to 
those  under  rectangular  beams. 

Case  I. — Beams    fixed   at   one    end    and 
loaded  at  the  other.    ^ 

S'  A'  d  =  W  *,  and  S"  A"  d  =  W  a;  (60) 


I 


- 


Case  II.  —  Beams   fixed  at   one   end  and 
loaded  uniformly. 


,  and  S"  A"  ct  =l/9 


(61) 


Case  III. — Beams  supported  at  both  ends 
and  loaded  at  some  intermediate  point. 


93 

(62) 

W.-£.«—  W(aj—  m)  6'  A'  d,  or,  =S"  A"  d 
I 

Case  IV.  —  Beams  supparted  at  both  ends 
and  loaded  uniformly. 


S'  A'rf=Vs  wx(l—  a;)  =  S"  A"  d 


Case  V.  —  A  single  moving  load  over  a 
beam  supported  at  both  ends. 

W    r 

8'  A'  d=  —.  -  (l—x)  —  S"  A"  d  (64 

i 

Case  VI.  —  A  distributed  moving  load 
may  be  considered  as  included  in  Case 
IV 


The  formulae  for  shearing  stress  are 
identical  with  those  in  rectangular  beams. 

The  principles  of  the  uniform  strength 
of  beams  may  be  applied  to  flanged  beams 
as  they  were  to  rectangular  beams.  The 
discussion  is  analogous  to  that  already 
-iven. 


MOMENT    OF    RESISTANCE    OF    BEAMS    DETER- 
MINED   GEOMETRICALLY. 

The  following  method  of  obtaining  the 
moment  of  resistance  of  beams  is  of  easy 
application,  and  in  many  cases  of  unsym- 
metrical  cross  section  is  the  simplest  that 
can  be  used : 

1.  For  illustration,  take  a  beam  of  rec- 
tangular cross  section.  Let  G  P  (Fig.  63) 
be  the  cross  section  at  some  point  of  this 
beam.  The  stresses  on  the  fibres,  as  we 
have  already  seen,  increase  just  in  propor- 
tion as  we  go  from  the  neutral  axis 
towards  the  upper  or  lower  surface  of  the 
beam,  and  may  for  any  vertical  slice  (as 
that  at  E  F)  be  represented  by  the  ordi- 
nates  of  two  triangles,  as  shown  in  Fig.  G4, 
where  E  I  (=F  J)  represents  the  stress  on 
the  outside  fibre.  For  the  cross  section  G 
P  (Fig.  03)  the  stresses  will  be  represented 
by  two  wedges,  the  bases  of  which  are  G 
M  and  M  R,  and  the  elevations  of  which 
are  the  triangles  shown  in  Fig.  64.  The 
volumes  of  these  wedges  give  the  amount 


95 


CO 

CO 


of  compressive  and  tensile  force  exerted  p.t 
the  cross  section  in  question,  arid  the 
points  in  G  P  under  the  centre  of  gravity 
of  the  wedges  give  the  "  centres  of  resist- 
ance," or  the  points  of  application  of  the 
resultants  of  these  forces. 


96 


As  a  geometrical  representation  of  the 
stresses  on  the  fibres,  these  wedges  are 
perfect,  for  the  perpendicular  ordinate  of 
the  wedge  gives  in  every  case  the  stress 
which  exists  in  the  fibre  over  which  it 
stands.  Thus  the  line  T'  V  (Fig.  64) 
represents  the  stress  on  each  fibre  in  the 
row  T  V  (Fig.  63). 

But  it  is  often  difficult  to  find  the  centre 
of  gravity  of  these  wedges  in  the  case  of 
curved  and  irregular  cross  sections,  and 
yet  this  must  be  done  before  we  can  know 
the  lever-arms  of  the  stresses.  To  render 
this  easier  to  do  we  may  represent  the 
stresses,  not  by  wedges,  but  by  prisms, 
the  centres  of  gravity  of  which  are  over 
the  centres  of  gravity  of  their  bases. 

Thus,  if  in  Fig.  65  we  draw  the  two 
shaded  tringles,  and  conceive  prisms  of  a 
height  =  E  I  (the  stress  on  the  outside 
fibre,  Fig.  64),  to  be  constructed  on  them 
as  bases,  we  shall  have  a  geometrical 
representation  of  the  stress  on  the  sec- 
tion G  P,  less  perfect  in  some  respects 


97 


FIG.  65. 

than  that  given  by  the  wedges  but  better 
suited  to  our  purpose. 

For,  note  that, 

1.  The  volume  of  the  prism  G  O  1J 
(Fig.  65)  is  equal  to  that  of  the  wedge 
G  M  (Fig.  63),  and  the  volume  of  any 
part  of  the  prism  cut  off  by  a  plane 
parallel  to  the  neutral  axis,  as  that  whose 
base  t'  O  vf  is  equal  in  volume  to  the 
corresponding  part  of  the  wedge  T  M, 
or  since  the  height  of  the  prism  is  con- 
stant, the  stress  on  the  surface  G  M  as 
we  go  out  from  the  neutral  axis  varies 


98 


as  the  area  of  the   triangle  which   forms 
the  base  of  the  prism. 

2.  The  vertical  slice  of  the  prism  stand- 
ing on  any  line  t  v'  represents  in  amount 
the  stress  on  the  line  of  fibres  t  r,  for  this 
stress  is  equal  to   the   corresponding   one 
in  the  wedge,  the  slice  of  the  prism  being 
as   much  higher  than  that  of  the   wedge 
as  t  v  exceeds  t'  v'.     Of  course  (except  in 
the  case  of  the  outside  fibres  in  the  row 
G  H)  each   ordinate    in  the  slice    of    the 
prism    no  longer  represents   the  stress  on 
the  fibre  over  which  it  stands,  as  was  the 
case  in  the  wedge. 

3.  The  moment  of  the    tensile    forces, 
for   instance,  will   equal   the   area  of  the 
prism  G   O   H   multiplied  by   its  height, 
(E  I  =  stress  on  outside  fibre  =  S).       The 
centre   of  resistance   of    these    forces,   or 
the  centre  of  gravity  of  the  prism  is  at  C 
(Fig.    65),   the   centre   of  gravity  of  the 
base   G   O    H.      The  triangle  G   O  H  is 
sometimes  called  the  ^effective  area"  of 
the  surface  G  M,  because  a  uniform  stress 
on   it    of    an    intensity  =  the    unit   stress 
at  G  H  gives  the  same  amount  of  resist- 


99 


FIG.  66. 

ance,  as  that  on  the  whole  area  G  M,  acted 
on  as  the  latter  is  by  a  varying  stress. 

Considering  the  stresses  represented  by 
the  two  prisms  whose  bases  are  G  O  II 
and  R  O  P  (Fig.  65),  as  concentrated  at 
the  centres  of  gravity  C  and  C'  (Fig.  67) 
of  these  bases,  and  taking  one  of  these 
points  as  C',  (Fig.  67)  as  the  centre  of 
moments,  we  have  in  the  case  represented 
in  the  figure: 

(Vol.  of  prism  G  O  H)xC  C'  =W  x 
or  if  b  —  breadth  and  d  =  depth  of  beam 


100 


FIG.  67. 


t 


(65) 


as  before. 

Corollary.  If  the  beam  be  square, 
b  —  dy  and 

M  =  |s&3  (66) 

II.  As  a  second  example,  take  a  square 
beam  so  placed  that  its  diagonal  will  be 
vertical.  Fig.  68  is  the  cross  section. 
Here  we  find  the  base  of  the  prism  of 
stress  by  points.  To  find  the  line  in  the 
base  of  the  prism  corresponding  to  the 


FIG.  68. 

stress  in  any  row  of  fibres,  such  as  A  B 
whose  distance  from  the  neutral  axis  is  O 
X,  proceed  as  follows: 

We  see  that  if  the  cross  section  were 
the  square  of  which  H  L  M  K  is  the 
half,  then  a'  b'  would  be  the  line  required, 
since  this  is  the  breadth  at  that  point  of 
the  triangle  H  O  K,  which  would  in  that 


102 

case  represent  the  base  of  the  prism  of 
stress.  Project  the  points  A  and  B  upon 
H  K.  Then  the  actual  row  (A  B)  of 
fibres  is  as  much  shorter  than  the  corres- 
ponding row  in  the  supposed  section  H 
M,  as  R  T  is  less  than  H  K,  and  conse- 
quently to  obtain  the  proper  line  in  the 
base  of  the  true  prism  of  stress,  a!  b'  must 
be  shortened  in  this  proportion.  Draw 
lines  from  R  and  T  to  O.  These  lines 
intersect  the  row  of  fibres  at  a  and  b. 
Then 


HK:  RT(  =  AB)::a'5':  ab         (67) 

Hence  a  b  is  the  line  required,  and  a  and 
b  are  two  points  in  the  outline  of  the  base 
of  the  prism  of  stress.  Any  number  of 
lines  as  n  v,  <fcc.,  may  be  gotten  similarly 
and  the  curve  drawn  through  the  points 
a  ..  n  ..  b  ..  v,  &c.,  will  give  the  form  of  the 
base  of  the  prism  of  stress.  This  base  is 
shaded  in  the  diagram. 

For  any  ordinate  of  the  curve  O  n  G,  as 
a  X,  we  have 

OX:  aX::  O  G:  RG  =  AX 


103 

•jt 

But   A  X  =  X  G   and    making   G  O  =  — 

and    putting    O  X  =  x    and  a  X  =y,  we 
have 


_  __  ....          (68) 

This  is  the  equation  of  a  parabola  with 
vertex  at  n,  half  way  between  II  K  and 
the  neutral  axis.  Hence  the  base  of  each 
prism  is  composed  of  parts  of  two  sym- 
metrical parabolas. 

Areas  of  the  bases.  Since  the  area  of  a, 
parabola  is  two-thirds  of  the  circumscrib- 
ing rectangle,  the  area  of  each  base 


But  n  v  =  y,R'    T'  and  R'  T  =Va  H  K 
.-.  nv  =  l/4  H  K  =  V4  <*' 

d'  d'       1  , 
/.  Area  =:  V,^.^=r/2 

The  centres  of  gravity  of  these  bases 
(and  consequently  of  the  prisms)  are  at  C 
and  C',  and  the  distance 


104 

Hence   the   moment   of   resistance  of   the 
fibres  about  C  or  C'  is 


(69) 

(S=height  of  prism  or  stress  on  external 
fibres  at  G  and  P.) 

Corollary.  To  compare  the  resistance 
of  the  beam  in  this  position  with  its  resist- 
ance when  lying  flat: 

Let  d=side  of  the  square  as  L  G. 
Then  d'  —d  fa  and  eq.  (69)  becomes 

M^S*V*=^S#'  (70) 

Comparing  this  with  eq.  (66),  we  see 
that  the  beam  offers  greater  resistance 
when  flat  in  the  proportion  of 

1        1 


In  solving  problems  with  diagonally 
placed  beams,  place  the  above  value  of  M 
equal  to  the  moment  of  the  weight  as  be- 
fore. 


105 

III.  Let  us  apply  this  method  to  a  T 
beam.  Take  for  example  the  cast-iron  j_ 
beam,  calculated  in  part  on  p.  .257  Ran- 
kine's  Civil  Engineering,  in  which  the  area 
of  the  flange  =  2/3  that  of  the  web.  As- 
sume the  flange  to  be  6  inches  by  1  inch, 
and  the  web  to  be  5  inches  by  .8  of  an  inch, 
and  draw  a  figure  of  the  cross  section  to 
scale  (Fig.  69). 

1.  As  the  top  and  bottom  of  this  section 
is  not  symmetrical,  it.  is  necessary  to  find 
the  position  of  the  neutral  axis,  which  is 
no  longer  at  the  half-depth.  This  may  be 
done  by  calculation,  or  by  a  simple  me- 
chanical process  as  follows: 

The  centre  of  gravity  of  the  cross  section, 
since  this  last  is  symmetrical  with  regard 
to  the  vertical  line  through  the  middle  of 
the  web,  must  lie  on  this  line.  Cut  accur- 
ately the  figure  of  the  cross  section  out  of 
card  board,  or  tin,  or  good  paper,  and 
suspend  it  freely  by  one  end  of  the  flange, 
suspending  also  from  the  same  point  a 
plummet,  Mark  the  line  of  the  plummet 
on  the  card  board,  and  the  centre  of  grav- 


<r 


ity  being  on  this  line  and  also  on  the 
middle  line  of  the  web,  will  be  at  their 
intersection  O.  By  measurement  this 
point  was  found  to  be  distant  from  A  B 
four  and  three-tenths  inches  (4.3"),  which 
is  also  the  value  by  calculations.  The 
line  L  M  drawn  through  this  point  is  the 
neutral  axis. 


*,        107 

2.  To  determine  the  bases  of  the  prisms 
of  stress.  On  the  upper  side  the  base  is 
the  triangle  O  A  B,  if  the  altitude  be  taken 
equal  to  the  stress  on  the  fibres  along  A  B. 
For  if  L'  M'  H  G  were  the  upper  half  sec- 
tion, O  G  H  would  be  the  base  of  the- 
prism  and  O  A  B  is  less  than  O  G  H,  in 
the  same  proportion  that  L  M  A  B,  the 
real  half  section,  is  less  than  L'  M'  II  G. 
Hence  (if  the  upper  be  the  compressed  side) 
the  total  compressive  force  is  equal  to  the 
prism  erected  on  A  O  B,  with  the  height 
equal  to  the  unit  stress  at  A  B. 

Below  the  axis  L  M. — For  convenience 
we  should  have  the  height  of  the  tension 
prism  equal  to  that  of  the  compression  one, 
and  the  base  must  be  determined  under 
this  condition.  Complete  the  large  rec- 
tangle G  H  Q  Y,  making  the  distance  of 
Q  Y  below  O=4.3  inches.  Draw  O  Q  and 
O  Y  and  the  shaded  trapezoid  J  N"  R  Z  cut 
out  on  the  flange  by  these  lines,  will  evi- 
dently be  the  portion  of  the  base  due  to 
the  flflnjrp.  Having  prolonged  the  lines  of 
the  web  to  T  and  V,  draw  O  T  and  O  V, 


108 

and  then  the  shaded  triangle  OKI  will  be 
that  part  of  the  base  due  to  the  portion  (L 
Z)  of  the  web  below  the  neutral  axis. 

The  total  tensile  and  compressive  forces 
being  always  equal,  and  the  height  of  the 
prisms  having  been  assumed,  each  =S'= 
stress  on  fibres  at  the  distance  of  A  B  from 
O,  the  bases  of  these  prisms  must  be  equal 
also.  This  necessary  equality  between  the 
area  of  O  AB  and  that  of  OKI  +  JNRZ, 
affords  a  means  of  testing  the  accuracy  of 
our  work  in  finding  the  position  of  O. 

3.     Area  of  base  GAB.     This  is, 


AB)  =  yf(4.3x.8)  = 
inches. 


4.  To  determine  the  distance  C  C'  (Fig. 
6D)  between  the  centres  of  gravity  of  the 
prisms,  which  distance  is  the  lever-arm  to 
to  be  used  when  one  of  these  points  is 
taken  as  the  centre  of  moments.  These 
centres  of  gravity  (C  and  C')  can  be  readily 
<UjUirmined  by  means  similar  to  those  em- 
ployed in  finding  the  centre  of  gravity 


t,     109 

of  the  cross-section  itself.  Thus,  cut  the 
shaded  areas  (Fig.  69)  out  of  card  board  of 
paper,  and  suspending  each  of  them  from 
two  points  in  sucession,  draw  vertical  lines 
through  the  points  of  suspension.  The  in- 
tersection of  these  two  lines  gives  the  cen- 
tre of  gravity.  In  the  present  case  they 
may  be  so  simply  obtained  by  calculation 
that  we  adopt  that  method.  The  centre 
of  gravity  of  A  O  B  is  =%  the  distance 
from  O  to  A  B  or  O  C  =%  (4.3)=  2.87  in- 
ches. As  to  the  shaded  part  below  O,  by 
using  the  ordinary  formula  for  the  centre 
of  gravity  and  taking  moments  around  O, 
we  find  the  distance 

OC'= 

j  0  N  R  x  %  (1.7)—  (O  J  K+Q I  Z)%(.7) ) 
|  ON  R—  2  (O  J  Kj 

2.2764— .01388 


3.0145— .2975 
Hence  the  distance 


=  1.318  inch. 


CC'=O  C+O  C'  =  2.87  +  1.318=4.18  ins. 
Hence,    since   S'   is   the   height   of   the 


110 

prisms,  the  moment  of  resistance  of  the  fi- 
bres is 

M=4.18xl.72S'  =  7.19S' 

If  it  be  desired  to  have  M,  not  in  terms 
of  S',  the  stress  along  A  B,  but  of  S"  the 
stress  on  the  lowermost  fibres  (at  F  P), 
we  have  since  the  stresses  increase  directly 
with  the  distance  from  O, 

S'   :S"::  I/ G  :  L' F  ::  4.3"  :  1.7" 
.  • .  S'=2.53.  S"  and  M=18.19.S" 

IV.  As  an  •  illustration  of  the  great  sav- 
ing of  labor  sometimes  effected  by  this 
process,  take  the  steel  rail  now  widely  used 
in  England,  the  cross- section  of  which  is 
given  to  scale  in  (Fig.  70).  The  determin- 
ation of  the  moment  here  by  calculation 
would  be  long  and  tedious.  The  dimen- 
sions of  the  cross  section  are  given  on  the 
figure. 

1.  The  centre  of  gravity  O  of  the  cross 
section  is  found  by  making  a  template  as 
in  the  last  case,  and  suspending  it  freely  by 
a  corner.  The  vertical  through  the  point 


FIG.  70. 

of  suspension  intersects  A  X  at  O,  which  is 
2.55  inches  below  A.  Through  this  point 
draw  L  M,  the  neutral  axis. 

2.  The  bases  of  the  prisms  of  stress  are 


determined    by    points    as     in     example 
II. 

Lay  off  OX=2.55  inches.  Draw  the 
rectangle  G  H  Q  Y.  Assume  the  height 
of  the  prisms  to  be  the  stress  in  the  fibre 
at  A.  Then  proceeding  as  in  example  II., 
the  line  of  the  base  correspending  to  any 
row  of  fibres,  as  B  D,  is  b  d.  Obtain  any 
number  of  points  in  the  same  way  as  b  and 
d,  and  through  these  points  draw  a  curve 
bounding  the  shaded  figure  A  b  Od.  Simi- 
larly below  the  neutral  axis,  t  v  is  the  line 
in  the  base  of  the  stress  prism  correspond- 
ing to  T  Y,  and  the  shaded  figure  O  N  R, 
is  'that  base  where  the  height  is  taken 
equal  to  the  unit  stress  at  A.  The  equality 
of  the  bases  in  area  is  the  test  of  accuracy. 

3.  To  determine  these  areas.  The  simplest 
plan  in  the  present  case  is  first  to  find  the 
area  of  the  cross  section  itself.  This  is  done 
as  follows:  The  rail  in  question  weighed 
84  Ibs.  per  yard,  and  the  steel,  of  which  it 
was  made  weighed,  .277  Ib.  per  cubic  inch. 


113 

Hence  if  A = area  of  cross  section  in  square 
inches 

(36  .  A).277=84.    .  • .  A=8.4  sq.  inches. 

Now  cut  out  of  the  same  card  board,  or 
paper,  templates  of  the  two  shaded  parts 
in  the  diagram,  and  also  of  the  cross  section 
itself,  and  weigh  them.  The  ratio  of  the 
weights  will  equal  that  of  the  areas. 

The  comparison  of  weights  may  be 
readily  made  by  the  means  of  a  suspended 
wire,  which  may  serve  as  a  temporary  bal- 
ance, the  templates  to  be  compared  being 
stuck  on  the  opposite  ends,  and  one  or  both 
moved  until  the  wire  is  evenly  balanced. 
The  weights  of  the  templates  being  in- 
vesely  as  their  distances  from  the  point  of 
suspension  of  the  wire,  their  areas  will  be 
in  the  same  proportion.  The  areas  of  the 
prisms  in  the  case  before  us  were  found  to 
be  equal  each  to  2.49  square  inches. 

4.  The  centres  of  gravity  of  these  bases 
are  found  in  the  same  way  as  that  of  the 
cross  section  itself.  The  point  C  was  thus 


114 

found  to  be  1.84  inches  above.  O  and  Cr 
to  be  1.66  inches  below  it.  Hence  the  dis- 
tance 

C  C'=3.5  inches. 

Therefore,  finally,  if  S'=unit  stress  at 
A,  the  moment  of  resistance  is, 


If  we  desire  M  in  terms  of  S"  (stress  at 
F  P,  we  have 

S'  :  S"  :  :  1^84  :  1,66 
.  *  .  S'  =  1.11S"     .  •  .M=9.67S" 

V.  A  circular  cross  section  (Fig.  71). 
Here  the  neutral  axis  of  course  =  L  M,  pass- 
ing through  the  centre.  Draw  the  circum- 
scribing rectangle  G  Y  and  obtain  the 
points  t,  by  v,  d,  &c.,  in  the  curve  bound- 
ing the  base,  as  heretofore.  Through  the 
points  so  found  draw  the  curves. 

Determine  the  areas  of  the  bases  of  the 
stress  prisms  by  comparing  them  with  the 
half-square  G  L  M  H.  Thus,  if  a  template  is 
cut  to  the  surface  of  one  of  these  bases  it 
will  just  equal  in  weight  the  template  cut 


115 


*'  x^x-x         ^- 

FIG.  71. 

to  the  surface  A  G  L  O  tb  A,  or,  in  other 
words,  the  shaded  surface  AbOdA  is  just 

(f 

one-third  of  the  rectangle  G  M,  or  l/s.-r  = 

</ 

»*•  '"'•',•  •':    • 

The  centres  of   gravity    C   and  C'  are 


116 

found   as   before  by   means  of  templates. 
In  this  way  it  was  found  that 

OC=OC'=.587  (O  A)  =.587.^ 

& 


The  height  of  the  prisms  being  S  (— 
stress  at  A  or  P)  the  moment  of  resistance 
is, 

M=S.-^(-.587  d\  =.0978  d*  S 
o 

(The  accurate  value  by  calculation  is 
M=.0982  d*  S.) 

Note.  —  The  curve  of  the  base  of  the 
prisms  is  a  lemniscate.  To  find  its  equation 
we  have  in  the  triangles  O  b  X'  and  O  B'A, 

bXf  :  AB'  (=BX')  ::  OX':  O  A 

And  taking  the  vertical  axis  as  that  of  X, 
and  the  horizontal  one,  as  that  of  Y,  the 
origin  being  at  O,  and  calling  the  co- 
ordinates of  the  circle  x'  and  y\  and  those 
of  the  lemniscate  x  and  y  we  have  : 


y:  y'::a:R,  or  y:  ^/E,2— x2 : :x :  R 
.  • .  R2  y2=  x  (R2— -a2). 


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MOTT  (H.  A.,  Jun.)  A  Practical  Treatise  on  Chemistry 
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MULLIN  (JOSEPH  P  ,  M.E.)  Modern  Moulding  and 
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NEWALL  (J.  W.)  Plain  Practical  Directions  for  Draw- 
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Useful  Hints  to  Gas  Consumers.  Second  edition, 
rewritten  and  enlarged.  By  Wm.  Paul  Gerhard, 
C.E. 

No.  112.— A  PRIMER  ON  THE  CALCULUS.     By  E.  Sherman 
Gould,  M.  Am.  Soc.  C.  E. 


J 


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and  for  Angles  to  Quarter  Degrees  between  1  degree 
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